document.write( "Question 312362: the perimeter of a rectangle is 80 meters. when the length is decreased by 7 meters and the width is increased by 7 meters the reulting figure is a square. find the dimensions of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #223503 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
the perimeter of a rectangle is 80 meters. when the length is decreased by 7
\n" ); document.write( " meters and the width is increased by 7 meters the reulting figure is a square.
\n" ); document.write( " find the dimensions of the rectangle.
\n" ); document.write( ":
\n" ); document.write( "\"the perimeter of a rectangle is 80 meters.\"
\n" ); document.write( "2L + 2W = 80
\n" ); document.write( "simplify, divide by 2:
\n" ); document.write( "L + W = 40
\n" ); document.write( "L = (40-W)
\n" ); document.write( ":
\n" ); document.write( "\"when the length is decreased by 7 meters and the width is increased by 7 meters the resulting figure is a square.\"
\n" ); document.write( "(L-7) = (W+7)
\n" ); document.write( "L = W + 7 + 7
\n" ); document.write( "L = W + 14
\n" ); document.write( "Replace L with (40-W)
\n" ); document.write( "40 - W = W + 14
\n" ); document.write( "40 - 14 = W + W
\n" ); document.write( "26 = 2W
\n" ); document.write( "W = \"26%2F2\"
\n" ); document.write( "W = 13 m is the width
\n" ); document.write( "then
\n" ); document.write( "L = 40 - 13
\n" ); document.write( "L = 27 m is the length
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check, adding 7 to 13 = 20 and subtracting 7 from 27 = 20 \n" ); document.write( "
\n" );